Inequalities represented graphicallyAQA A-Level Maths: Revision notes

The graph of $y=mx+c$ is a line which splits the plane into two regions. An inequality such as $y>x+1$ is satisfied by every point above the line $y=x+1$, and $y<x+1$ by every point below it. For a vertical line, $x>a$ is the region to the right of $x=a$ and $x<a$ is the region to the left; for a horizontal line $y>b$ is above and $y<b$ below. The boundary line is drawn dashed for a strict inequality ($<$ or $>$), because points on it are not in the region, and solid for $\le$ or $\ge$, because points on it are included. If a line is given as $ax+by\le c$, rearrange to make $y$ the subject, but remember to reverse the sign if you divide by a negative number.

If you are unsure which side to shade, use a test point. Pick a simple point not on the line, substitute it, and see whether the inequality is true. For $y<5-x$ use $(0,0)$: $0<5$ is true, so the side containing the origin is the required region. If the line passes through the origin, such as $y=3x$, use a point like $(1,0)$ instead: $0<3$, so $y<3x$ is the region below the line. Always state which region you have found. Some questions ask you to shade the required region $R$, others to shade the unwanted region so that $R$ is left clear. Read the instruction and label $R$.

A region defined by several inequalities satisfies all of them at once, so it is the intersection (overlap) of the individual regions. Draw each boundary line with the correct style, decide the side of each, and find the overlap. The corners of the region are found by solving the equations of two boundary lines simultaneously. Example: $y>x+1$ and $y\le5-x$. Solving $x+1=5-x$ gives the vertex $(2,3)$; the region lies above the dashed line $y=x+1$ and on or below the solid line $y=5-x$. The point $(1,3)$ is in $R$ because $3>2$ and $3\le4$, but the vertex $(2,3)$ is not, because the inequality $y>x+1$ is strict. To count whole-number points in a region, go through each integer $x$ and count the integer $y$ values allowed, remembering to include a boundary only if its line is solid. For $x\ge1$, $y\ge2$, $2x+y\le10$ there are $7+5+3+1=16$ such points.

A quadratic inequality in $x$ and $y$ such as $y>ax^2+bx+c$ is represented in the same way. Sketch the boundary curve $y=ax^2+bx+c$ (dashed for $>$ or $<$, solid for $\ge$ or $\le$), then take the points above the curve for $y>\ldots$ and below it for $y<\ldots$. For $y>x^2-4x+3$, the region is the points above the $\cup$-shaped curve, which is the 'inside' of the bowl. For $y<x^2-4x+3$ it is the points under the curve, which is the 'outside'. A region between a curve and a line is defined by two inequalities. For example, 'above $C$ and below $L$' for $C:y=x^2-4x+3$ and $L:y=x-1$ is $y>x^2-4x+3$ and $y<x-1$. The curve and line meet where $x^2-4x+3=x-1$, which gives $x=1$ and $x=4$, and the region exists only between these $x$-values.

An inequality in one variable can be solved from a graph. To solve $f(x)>0$, find where the graph of $y=f(x)$ is above the $x$-axis; for $f(x)<0$, where it is below the axis. The roots are the end points. Example: $y=x^2-2x-8=(x-4)(x+2)$ is $\cup$-shaped, crossing the $x$-axis at $-2$ and $4$ with minimum point $(1,-9)$ and $y$-intercept $-8$. It is below the axis for $-2<x<4$, so $x^2-2x-8<0$ for $-2<x<4$; and above for $x<-2$ or $x>4$. To solve $f(x)>g(x)$, sketch both graphs and find where the first is above the second. The solutions are bounded by the $x$-coordinates of the intersection points. Example: $x^2-2x-8<2x-3$ becomes $x^2-4x-5<0$, so $C$ is below $L$ for $-1<x<5$. Give the answer using 'and' for one piece and 'or' for two pieces.